A tangent PT is drawn to the circle `x^2 + y^2= 4` at the point `P(sqrt3,1)`. A straight line L is perpendicular to PT is a tangent to the circle `(x-3)^2 + y^2 = 1`Common tangent of two circle is: (A) `x=4` (B) `y=2` (C) `x+(sqrt3)y=4` (D) `x+2(sqrt2)y=6`A. `x-sqrt3y=1`B. `x+sqrt3y=1`C. `x-sqrt3y=-1`D. `x+sqrt3y=5`

A tangent PT is drawn to the circle `x^2 + y^2= 4` at the point `P(sqr

1.	A tangent PT is drawn to the circle `x^2 + y^2= 4` at the point `P(sqrt3,1)`. A straight line L is perpendicular to PT is a tangent to the circle `(x-3)^2 + y^2 = 1`Common tangent of two circle is: (A) `x=4` (B) `y=2` (C) `x+(sqrt3)y=4` (D) `x+2(sqrt2)y=6`A. `x-sqrt3y=1`B. `x+sqrt3y=1`C. `x-sqrt3y=-1`D. `x+sqrt3y=5`
Answer» Here, tangent to `x^(2)+y^(2)=4 "at" (sqrt3,1) "is" sqrt3x+y=4" "...(i)` AS, L is perpendicular to `sqrt3x + y = 4` `rArrx-sqrt3y=lambda` which is tangent to `(x-3)^(2)+y^(2)=1` `rArr (\|3-0-lambda\|)/(sqrt(1+3))=1` `rArr \|3-lambda\|=2` `rArr 3-lambda=2,-2` `therefore lambda=1,5` `rArrL:x-sqrt3y=1, x=-sqrt3y=5`

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